BIOCHEMISTRY 1. Consider the equilibrium between the boat forms of glucopyranose

Question

BIOCHEMISTRY

1.Consider the equilibrium between the boat forms of glucopyranose (both alpha and beta forms combined) and the chair forms of glucopyranose (both alpha and beta forms combined) as follows: boat <--> chair The chair form is 5.98 kcal/mol lower in energy than the boat form. Given the above information, what is the ratio of chair-form molecules to boat-form molecules? Assume a temperature of 25 degrees C (298 K). Report your answer to the nearest ones.

2.Consider the equilibrium between the pyranose forms of glucose (both and forms combined) and the furanose forms of glucose (both and forms combined) as follows: glucopyranose <--> glucofuranose The furanose forms are approximately 3.5 kcal/mol higher in energy than the pyranose forms. Given this information, what is the ratio of pyranose-form molecules to furanose-form molecules? Assume a temperature of 25 °C (298 K). Report your answer to the nearest ones.

3.Glucose dissolved in solution will contain 0.02% of the linear form of the molecule. Given this information, how much higher (or lower) in energy (in terms of kcal/mol) is the linear form of glucose compared to all of the cyclic forms of glucose (mentioned in "Chair Form vs. Boat Form" and "Pyranose Form vs. Furanose Form")? Assume a temperature of 25 °C (298 K). Report your answer to the nearest tenth and use a negative sign if you think the linear form is lower in energy.

4.-D-glucopyranose is 0.34 kcal/mol higher in energy than -D-glucopyranose. Given this information, what percent of glucose molecules are in the -D-glucopyranose form? Hint: given your answers to the previous "one form vs. another form" problems, do you need to consider the linear and/or furanose forms, or can you assume that essentially all of the glucose is partitioned between the -D-glucopyranose form and the -D-glucopyranose form? Report your answer as a percentage to the nearest ones.

Explanation / Answer

Given that; The chair form is 5.98 kcal/mol lower in energy than the boat form.

Assume if boat form has energy 10 Kcal /mol then chair has 10.0-5.98= 4.02 kJ/mol kcal/mol

Now we determine K:
K = e^(dG / (-8.315 x 10^-3 kJ/degxmol X 298 K))
or
K = e^(dG / -2.478kJ/mol)

For chair form:
K = e^(4.02 kJ/mol / -2.478kJ/mol)

K=0.198

For BOAT form:
K = e^(10.0 kJ/mol / -2.478kJ/mol)

K=0.0176

Here K = products / reactants.

The chair form (more stable) ---- > boat form(less stable)

The more stable conformation can therefore be thought of as the reactant and the less stable as the product.

K = product / reactant
0.0176 = 0.0176 / 1

To find the ratio of chair-form molecules to boat-form molecules, convert it to a percentage in relation to 1 by:

1 / (1 + 0.0176) = 0.983 (for the stable conformation).

the ratio of chair-form molecules to boat-form molecules:0.983/0.0176

=55.85

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